Showing How to Imply Proving The Riemann Hypothesis

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1 EUROPEAN JOURNAL OF MATHEMATICAL SCIENCES Vol., No., 3, 6-39 ISSN Showing How to Imly Proving The Riemann Hyothesis Hao-cong Wu A Member of China Maths On Line, P.R. China Abstract. Using the integral convergence attaching analytic roerties rove the Riemann hyothesis. Mathematics Subject Classifications: M6 Key Words and Phrases: Lalace Transform, Converges Uniformly, Analytic.. Introduction We show how the function roerties rove the integral x 3 +ɛ d x converges absolutely and uniformly for ɛ>, which concludes the integral d x x s+ is analytic for Re(s)>, thus this concludes the roof of the form for ɛ>. ψ(x)= x+ O(x +ɛ ) addresses: whc8778@63.com (H. Wu) htt:// 6 c 3 EJMATHSCI All rights reserved.

2 H. Wu/Eur. J. Math. Sci., (3), Preliminaries.. Basic Proerties of The Zeta Function For Re(s)> we know that the zeta function defined by the series and the Euler roduct namely ζ(s)= n s n= ( s ), n s= ( s ). n= The series converges absolutely and uniformly for Re(s) +δ, with anyδ>, and the roduct converges absolutely and uniformly for Re(s) +δ, with anyδ>. The reresentation of the zeta function as such a roduct shows thatζ(s) for Re(s)>. In fact,we often obtain the following identities,valid for Re(s)>: and ζ(s)= n s = s n= s n= n d x x s+= s [x] s d x= xs+ s s d x x s+= s [x] d x, xs+ n x {x} d x. xs+ The symbol[x] denotes the greatest integer x, it is called the integral art of x, the number {x}= x [x] is called the fractional art of x, it satisfies the inequalities {x}<, with {x}= if and only if x is an integer. We know the integral {x} x s+ d x converges absolutely, and uniformly for Re(s) δ, with anyδ>. Also, for Re(s)> we know the identity (ii): φ(s)= ζ (s) ζ(s) = n= Λ(n) log n s = s = s ψ(x) s d x= xs+ s + s x s+ d x, where the sum is extended over all rimes. log s

3 H. Wu/Eur. J. Math. Sci., (3), We define ψ(x)= m x log = Λ(n). TheΛ(n) is log when m such that n= m, or otherwise it is. Where the sum is taken over those integers of the form m that are less than or equal to x. Here is a rime number and m is a ositive integer.we also see thatζ(+ it),ζ(it), and when the function ζ /ζ(s) has no oles on the region >Re(s)>, then which imlies that the functionζ(s) has no zeros on the region >Re(s)>. We intimately know the result for Re(s)>, and the identity (ii): We define ζ /ζ(s)= φ(s)= ζ /ζ(s)= s s + s n x log s d x for Re(s)>. x s+ log Φ(s)= s for Re(s)>. Here is a rime number. The sum definingφ(s) converges uniformly and absolutely for Re(s) +δ, by the same argument as for the sum defining the zeta function. We merely use the fact that givenɛ>, Since log n n ɛ for all n n (ɛ). e x = + x+...+ x n it follows that for ositive x and for all n, we have x n e x > n! + x n+ (n+)! +..., x (n+)!. For any fixed n the right-hand side tends to infinity as x, it follows that e x grows faster than any fixed ower of x. We can write x n = o(e x ) to mean x n lim x e x = for all n. We see that log x= o(x δ ),δ>. In other words log x grows slower than any fixed ositive ower n of x.

4 H. Wu/Eur. J. Math. Sci., (3), Basic Results In a way, we need go through the basic definitions with their results concerning functions, and we omit some definitions and roofs which are just elementary knowledge such as: We recall the notation: f = O(g) mean that f, g are two functions of a variable x defined for all sufficiently large, and g is ositive, there exists a constant C> such that f(x) C g(x) for all x sufficiently large. In fact,we haveψ(x)= x+ O(x λ ) for λ<, andψ(x)=o(x). We also recall the results, letρ be all nontrivial zeros ofζ(s), we know ψ(x)= x+ O(x sure(ρ)+ɛ ) for everyɛ>. Intimately, we can getλ su Re(s)+ɛ by the result thatζ(s) has no zero on the region Re(s)>λ. Using a basic fact from roerties of the zeta function that ζ(s) certainly has an infinite number of the nontrivial zeros in the region Re(s) and the symmetry of the zeros such that we haveλ. We recall the definitions:let U be a subset of the comlex lane, we say that U is oen set if for every oint z in U, there is a disc D(z, r) of radius centered at z such that this disc D(z, r) is contained in U. And we know a set what is called closed,what is called bounded,what is called continuous,comact.we also know a series of sequence and a series of functions what is said to converge uniformly, converge absolutely. We have the usual tests for comact sets, and some for convergence, etc, those definitions with their results..3. Some Theorems We make the recise theorems omitted its roof as follows[cf. ]. Theorem. A set of comlex numbers is comact if and only if it is closed and bounded. Theorem. Let S be a comact set of comlex numbers,and let f be a continuous function on S. Then the image of f is comact. Theorem 3. Let S be a comact set of comlex numbers, and let f be a continuous function on S. Then f is uniformly continuous, i.e. givenɛ there existsδsuch that whenever z, w S and z w <δ, then f(z) f(w) <ɛ. Theorem 4. Letγbe a ath in an oen set U and let g be a continuous function onγ(i.e. on the imageγ([a, b]) ifγis defined on[a, b]). If z is not onγ, define f(z)= γ g(ζ) ζ z dζ. Then f is analytic on the comlement ofγin U, and its derivatives are given by[cf.,.3-3] f (n) g(ζ) (z)= n! dζ. (ζ z) n+ γ

5 H. Wu/Eur. J. Math. Sci., (3), Let f be a continuous function on the real numbers and assume that there is constants A, B such that f(t) Ae Bt for all t sufficiently large. Moreover,and assume for simlicity that f is bounded, iecewise continuous, whence B. What we rove will hold under much less restrictive conditions: instead of iecewise continuous,it would suffice to assume that the integral b f(t) dt a exists for every air of numbers a, b. We shall associate to f the Lalace transform g defined by g(z)= f(t)e zt dt for Re(z)>, we can then aly Lemma (the differentiation lemma) of subsection 4., whose roof alies to a function satisfying our conditions (iecewise continuous and bounded), and then we easily conclude that g is analytic for Re(z)>. 3. The Three Basic Theorems Theorem 5. The functionφis meromorhic for Re(s)>. Furthermore, for Re(s), we have ζ(s) and Φ(s) s has no oles for Re(s). Proof. For Re(s)> the Euler roduct shows thatζ(s). We know the result for Re(s)>, and the identity (ii): ζ /ζ(s)= log s φ(s)= ζ /ζ(s)= s s + s x s+ d x. Using the geometric series we get the exansion s = s = s + s+ s+... s So the identity (ii) it satisfies = s+ s+... () ζ /ζ(s)=φ(s)+ h (s), where h (s) log B for some constant B. () s

6 H. Wu/Eur. J. Math. Sci., (3), But the series log n converges absolutely and uniformly for Re(s) +δ, with anyδ>, and we know the intimate result that the functionζ(s) extends to an analytic function on the region s Re(s)>, this result and (ii) imly thatφis meromorhic for Re(s)>, and has a ole at s= and at the zeros ofζ, but no other oles in this region. We omit the roof thatζhas no zero on the line Re(s)=, which is just the intimate result. We shall now rove secial cases of the following theorems concerning differentiation under the integral sign which are sufficient for our alications. We first rove a general theorem that the uniform limit of analytic functions is analytic. This will allow us to define analytic functions by uniformly convergent series. Theorem 6. Let f n be a sequence of analytic functions on an oen set U. Assume that for each comact subset Kof U the sequence converges uniformly on K, and let the limit function be f, i.e. lim f n = f. Then f is analytic. Proof. z U, and let D R be a closed disc of radius R centered at in z and contained in U. Then the sequence f n converges uniformly on D R. Let C R be the circle which is the boundary of D R. Let D R be the closed disc of radius R centered at z. Then for z D R we have n s f n (z)= f n (ζ) πi ζ z dζ, C R and ζ z R. Since f n converges uniformly to f(z), forz z R, we get f(z)= πi C R f(ζ) ζ z dζ. By Theorem 4 it follows that f is analytic on a neighborhood of z. Since this is true for every z in U, we have roved what we wanted. Theorem 7. Let f n be a sequence of analytic functions on an oen set U converging uniformly on every comact subset K of U to a function f. Then the sequence of derivatives f n converges uniformly on every comact subset K, and lim f n = f. Proof. Cauchy s formula exresses the derivative f n as an integral, cover the comact set with a finite number of closed discs contained in U, and sufficiently small radius. And one can argue as in the revious Theorem 6. For the sake of comleteness we shortly state the similar result,and we have f n (z)= πi C R f n (ζ) ζ z dζ,

7 H. Wu/Eur. J. Math. Sci., (3), and ζ z R. Since f n converges uniformly to f(z), i.e. lim fn = f, for z z R, we get f(z)= πi C R f(ζ) ζ z dζ. By Theorem 4, we obtain a bound for the derivative of an analytic function in term of the function itself, we see that f n (z)= f n (ζ) πi C R (ζ z) dζ, and and Therefore f (z)= πi C R lim f n = f. lim f n = f. f(ζ) (ζ z) dζ, 4.. Stating Some Basic Conditions We know that the form 4. The Three Main Lemmas ψ(x)= x+ O(x +ɛ ) for ɛ>imlies the Riemann Hyothesis.We shall now rove the main lemmas, which constitute the delicate art of the roof. We shall also deal with integrals deending on a arameter. This means a function f of two variables, f(t,z), where z in some domain U in the comlex numbers, and we let D f and D f be the artial derivatives of f with resect to the first and second variable resectively. The integral f(t,z)dt= lim B B f(t,z)dt is said to be uniformly convergent for z U if, givenɛ, there exists B such that B < B < B, then B f(t,z)dt <ɛ. B The integral is absolutely and uniformly convergent for z U if the same condition holds with f(t,z) relaced by the absolute value f(t,z).

8 H. Wu/Eur. J. Math. Sci., (3), The Three Lemmas Lemma (The Differention Lemma). Let I be an interval of real numbers, ossibly infinite. Let U be an oen set of comlex numbers.let f= f(t,z) be a continuous function on I U. Assume: (i) For each comact subset K of U, the integral is uniformly convergent for z K. (ii) For each t the function z f(t,z) is analytic. Let F(z)= I I f(t,z)dt f(t,z)dt. Then F is analytic on U, D f(t,z) satisfies the same assumtions as f, and F (z)= I D f(t,z)dt. Proof. Let I n be a sequence of finite closed interval, increasing to I. Let D be a disc in the z-lane whose closure is contained in U. Letγbe the circle bounding D. Then for each z in D, we have f(t,z)= f(t,ζ) πi ζ z dζ, γ So F(z)= f(t,ζ) πi ζ z dζdt. I γ Ifγhas radius R center z, consider only z such that z z R, then ζ z R. For each n we can define F n (z)= f(t,ζ) πi ζ z dζdt. I n γ In view of the restriction on z above, we may interchange the integrals and get Cauchy s formula exresses the F n as an integral, covering the comact set with a finite number of closed discs contained in U, and sufficiently small radius, i.e. F n (z)= πi γ f(t,ζ)dt dζ. ζ z I n

9 H. Wu/Eur. J. Math. Sci., (3), We can aly our conditions that the integral I f(t,z)dt is uniformly convergent and f= f(t,z) is uniformly continuous on the comact subset I K of I U, we obviously get that I f(t,z)dt is continuous on the comact subset K of U on theγ. Hence I n f(t,z)dt is continuous on the athγ. Then F n is analytic by Theorem 4. By Theorem 6 and Theorem 7 with our assumtion,the integrals over I n converge uniformly to the integral over I. Hence F is analytic,being the uniform limit of the function F n for z z R. On the other hand, F n (z) is obtained by Theorem 7, and converges uniformly to F (z), i.e. the uniform limit of F n (z) is: This roves the theorem. lim F n (z)= F (z)= I D f(t,z)dt. Lemma. Let f be bounded, iecewise continuous on the real numbers. Let f the Lalace transform g defined by g(z)= f(t)e zt dt for Re(z)>, then g is analytic in the region Re(z)>. If g extends to an analytic function for Re(z), then exists and is equal to g(). f(t)dt Proof. Let B be a bound for f(t), that is f(t) B for all t, then we can aly the differentiation lemma to conclude that g is analytic in the region Re(z)>, we omit its roof which is just a intimate result. For T> define g T (z)= T f(t)e zt dt. Then g T is an entire function,as follows at once by the differentiation lemma. We have to show that lim T g T()= g().

10 H. Wu/Eur. J. Math. Sci., (3), Letδ>and let C be the ath consisting of the line segment Re(z)= δ and the arc of circle z =R and Re(z) δ, as shown on the Figure. Figure : Line segment Re(z)= δ and the arc of circle z =R and Re(z) δ. By our assumtion that g extends to an analytic function for Re(z), we can takeδ small enough so that g is analytic on the region bounded by C, and on its boundary. Then g() g T ()= (g(z) g T (z))e Tz + z dz πi C R z = H T (z)dz, πi C where H T (z) abbreviates the exression under the integral sign. Let B be a bound for f(t), that is f(t) B for all t. Let C + be the semicircle z =R, and Re(z). Then H T (z)dz πi B R. (3) C + First note that for Re(z)> we have g(z) gt (z) = f(t)e zt dt B T T B e zt dt= Re(z) e Re(z)T ;

11 H. Wu/Eur. J. Math. Sci., (3), and for z =R, etz (+ z R ) z = R ere(z)t z + z R R = ere(z)t Re(z) R. Taking the roduct of the last two estimates and multilying by the length of the semicircle gives a bound for the integral over the semicircle,and roves the claim. Let C be the art of the ath C with Re(z)<. We wish to estimate g(z) gt (z) e Tz + z dz πi R z. C Now we estimate searately the exression under the integral with g and g T. We have g T (z)e Tz + z dz πi C R z B R. (4) Let S be the semicircle with z =R and Re(z)<. Since g T is entire, we can relace C by S in the integral without changing the value of the integral, because the integrand has no oles to the left of the y-axis. Now we estimate the exression under the integral sign on S. We have gt (z) T = f(t)e zt dt B T e Re(z)t Be Re(z)T Re(z). For the other factor we use the same estimate as reviously. We take the roduct of the two estimates,and multily by the length of the semicircle to give the desired bound in(4). Third, we claim that C g(z)e Tz + z R dz, (5) z as T. We can write the exression under the integral sign as g(z)e Tz + z R z = h(z)etz, where h(z) is indeendent of T Given any comact subset K of the region defined by Re(z)<, we note that e Tz raidly uniformly for z K, as T. The word "raidly" means that the exression divided by any ower T N also tends to uniformly for z in K, as T. From this our claim(5) follows easily. We may now rove the Lemma. We have f(t)dt= lim T g T (),

12 H. Wu/Eur. J. Math. Sci., (3), if this limit exists. But givenɛ, ick R so large that B <ɛ. Then by(5), ick T so large that R g(z)e Tz + z dz C R z <ɛ. Then by(3),(4) and(5) we get g() gt () < 3ɛ. This roves Lemma. We claim that it also concludes the integral x 3 +ɛ d x is convergent for ɛ>. Observe that the function ψ is iecewise continuous. In fact, it is locally constant: there is no change inψbetween rime numbers. The alication of Lemma is to rove: Lemma 3. The air of integrals and converge for ɛ>. x 3 +ɛ d x, x 3 +ɛ d x and Proof. Let f (t)= ψ(et ) e t e ( (λ λ))t, f (t)= ψ(e t ) e t e ( (λ λ))t, whereλsatisfies the conditionsψ(x)= x+ O(x λ ), λ<, givenλ λ ɛ for ɛ>, show that given anyλ λ ɛ satisfies (λ λ) 3 +ɛ for ɛ> including the case (λ λ)= 3 +ɛ, and whereλ is deendent ofɛ,λcan be indeendent ofɛ. Then f i is certainly iecewise continuous, and is bounded by the formulasψ(x)=o(x) andψ(x)= x+ O(x λ ) for λ<, where i=,. Making the substitution x= et in the desired integral, d x= e t dt, we see that and x 3 +ɛ d x= x 3 +ɛ d x= f (t)dt, f (t)dt.

13 H. Wu/Eur. J. Math. Sci., (3), Therefore these suffice to rove that the integrals on the right converge, it suffices to rove that the Lalace transform g i of f i is analytic for Re(z), so we have to comute the Lalace transform g i. We claim that in the case i=, g (z)= φ z+ (λ λ) z+ (λ λ) z (λ λ). Once we have roved the formula, we can then aly Theorem 5 and the identity (ii) to conclude that g (z) is analytic for Re(z), thus concluding the art of the roof of Lemma 3. Now to comute g i (z) when i=, we use the identity (ii) we obtain and φ(s) s s = x s+ d x, g (z)= φ z+ (λ λ) z+ (λ λ) z (λ λ) = = ψ(e t ) e t e (z+ (λ λ))t et dt= f (t)e zt dt x z+ (λ λ) d x where Re z+ (λ λ) >, and so Re(z)>λ λ, we haveλ λ ɛ for ɛ>, which by Theorem 5 and the identity (ii) show that g (z) is analytic for Re(z)>λ λ, and then it is analytic for Re(z). This gives us the Lalace transform of f and concludes the art of the roof of Lemma 3. We claim that in the case i=, ψ(e t ) e t g (z)= e (z+ (λ λ))t et dt= x d x= f z+ (λ (t)e zt dt, λ) where ψ(e t ) e t f (t)= e, ( (λ λ))t whence the function converges absolutely for Re(z) and uniformly for Re(z) δwith δ from the basic fact that the integral converges absolutely and uniformly for n + k, with any k >. Given any d x x n λ λ ɛ for ɛ>, with ψ(x)= x+ O(x λ ), λ<.

14 REFERENCES 39 When Re(z), we have Re(z)+ λ>, Re(z)+ λ >. Then by Lemma (the differentiation lemma) which show that g i (z) is analytic for Re(z) when i=. This gives us the Lalace transform of f and concludes the art of the roof of Lemma3. Hence this roves the lemma. 5. Conclusions Using Lemma 3 concludes that the integral d x x s+ converges absolutely and uniformly for Re(s)>, and by Lemma (the differentiation lemma) concludes the integral d x x s+ is analytic for Re(s)>, which immediately follows that the functionζ /ζ(s) has no oles on the region >Re(s)> from the formula (ii), and which imlies that the functionζ(s) has no zeros on the region >Re(s)>. Thus concluding the roof of the formψ(x)= x+ O(x +ɛ ) for ɛ >, and it also concludes the roof of the Riemann Hyothesis. ACKNOWLEDGEMENTS The author would like to thank Professor Maohua Le for his invaluable comments and encouragements during his stay in Zhanjiang Normal College, and esecially wants to thank the readers of Euroean Journal of Mathematical Sciences, for making the journal successful. Finally, the author thanks the editorial team of EJMS. References [] S. Lang. Comlex Analysis. Sringer-Verlag and Beijing World Publishing Cororation, 4th Edition, Beijing. 3.